Question

XXIII. An isotonic solution will produce an osmotic pressure of 7.84 atm measured against pure water at human temperature (37.0 °C). How many grams of sodium chloride must be dissolved in 500 mL of water to produce an isotopic solution? Na = 23.00; CI = 35.45 A) 9.01 g B) 4.51 g C) 18.0 g D) none of them

Answer 1

The correct answer is - (B.) 4.51 g

The formula to calculate osmotic pressure is-

PV= inRT or PV= i$\times$(w/M)$\times$RT

where,

P= osmotic pressure= 7.84atm (given)

V= volume of solvent= 500mL (given) = 500/1000 L= 0.5L

i= van't hoff factor=2 for NaCl (as NaCl dissociates into Na+ and Cl- ions, i.e two ions, so i=2)

R= gas constant= 0.0821 Ltrs atm K^-1 Mol^-1

T= temperature = 37degrees C (given)= (37+273) K= 310K

n= number of moles = w/M

w= gram weight of solute ( to be calculated)

M= molar mass of solute = (no.of atoms $\times$ atomic weight of each element), for NaCl it is- (1$\times$atomic weight of Na+1$\times$atomic weight of Cl ) = [(1$\times$ 23) + (1$\times$35.45)]= [23+35.45] = 58.45 g/mol (atomic weights- given)

Now, let's plug these values in the eqn-

PV= i$\times$(w/M)$\times$RT

7.84atm $\times$ 0.5L = 2$\times$(w/58.45 g/mol) $\times$ 0.0821 L atm K^-1 Mol^-1$\times$ 310K

On re- arranging,

w= (7.84atm $\times$ 0.5L $\times$ 58.45 g mol^-1) / (2$\times$0.0821 L atm K^-1 Mol^-1$\times$ 310K)

On cancelling some units

w= (7.84$\times$ 0.5$\times$ 58.45 g) / (2$\times$0.082$\times$ 310)

w= (229.124 g) / (50.84)

w= 4.506 g , which can be rounded off as 4.51 g

Hence, 4.51 grams of NaCl must be dissolved in 500mL of water to produce an isotopic solution.